# What are the natural numbers x if y is integer? y=(6x-8)/(2x+1)

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Use cross multiple:

y(2x+1) = 6x-8

2yx +y = 6x - 8

2yx - 6x = -y-8

2x(y-3) = -(y+8)

x = -(y+8)/2(y-3)

If y = 2, x = 5

This is the only one meeting the requirement.

We'll write the numerator: 6x-8=2x+4x -9+1

y=(6x-8)/(2x+1)=[(2x+1)/(2x+1)]+[(4x-9)/(2x+1)]

y=1+[(4x-9)/(2x+1)]

We'll try to do the same with the ratio [(4x-9)/(2x+1)]=[(2x+1+2x-1-9)/(2x+1)]=

[(2x+1)/(2x+1)]+[(2x-10)/(2x+1)]=1+[(2x-10)/(2x+1)]

So y = 1+1+[(2x-10)/(2x+1)]

We'll follow the same steps:

[(2x-10)/(2x+1)].

[(2x-10)/(2x+1)]= [(2x+1-1-10)/(2x+1)]=1- [11/(2x+1)]

y=1+1+1- [11/(2x+1)]

y=3- [11/(2x+1)]

If y is integer, the fraction [11/(2x+1)] has to be also an integer number. For this reason, (2x+1) has to be the divisor of the number 11. So, (2x+1) could be:+1,-1,+11,-11.

Now, we'll put (2x+1) = 1

2x=0, x=0 and is a natural number, so it follows the constraint

y=(6*0-8)/(2*0+1)

y=-8/1

y=-8

(2x+1) = -1

2x=-2

x=-1, but "-1" is not a natural number, so (2x+1) is different from -1

(2x+1) = 11

2x=10

x=5 and is a natural number.

y=(6*5-8)/(2*5+1)

y=22/11

y=2

(2x+1) = -11

2x=-12

x=-6 and is not a natural number.

**The natural values of x, that makes the ratio y integer, are: {0 ; 5}.**